A few minutes ago, Marijn wrote:
>> And so it does, but even in the simplified example I find it hard to
> reason about it and come to a conclusion on whether it should return
> 1 or 2.
Yes, it is a little tricky.
> Do you just look at:
>> (define-syntax-rule (make-variable __E__)
> (lambda () (let ([_id_ 1]) __E__)))
>> and say "__E__ and _id_ are not introduced by the same syntax-rule
> pattern, so by hygiene _id_ is not free in __E__ and thus (let ((_id_
> 1)) ...) cannot bind anything in __E__" or how does it work?
This sounds like a valid intuition.
(BTW, if it was't clear, the above could be fixed by making `_id_' an
input to the `make-variable' macro, which should be easy to do in your
original code too.)
--
((lambda (x) (x x)) (lambda (x) (x x))) Eli Barzilay:
http://barzilay.org/ Maze is Life!